Question

Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.

• A.
• B.
• C.
• D. none of these

Answer 1

The correct option is C

The spring will be stretched to balance the tension in the string.

Let the elongation be $x_0$

$\Rightarrow$ K$x_0$ = $T_0$

$\Rightarrow$ 2 $T_0$ = mg

$\Rightarrow$ 2 K$x_0$ = mg ....(1)

Now if I pull the pulley by x downwards then there is an extra AB = C D = X part in the string.

$\Rightarrow$ a total of 2x elongation. Since the string is non stretchable so obviously spring is further elongated by 2x on top of the $x_0$ at equilibrium. So Now the spring force is K (2 x + $x_0$) which will be equal to the tension in the strings.

K(2x + $x_0$) = T ---------------(ii)

$\Rightarrow$ 2T - mg = ma

2k(2x + $x_0$) - 2K$x_0$ = ma

from equation (i) & (ii)

a = $\frac{4kx}{m}$